This is trivial, isn't it? Let the number of digits of initial number is $m$, then the possible number on the blackboard is below $9^m$.So by PHP, some natural number which is below $9^m$ appears infinitely many times.

Wait I think the interpretation of the problem is this: Let P(m) denote the product of the nonzero digits of m, and $m_i$ = $P(m_{i-1})$ + $m_{i-1}$, where n = $m_0$. I think you're trying to prove that over all i $\in \mathbb{N}$, there's some infinitely long sequence $a_1, a_2, \dots$ such that $P(m_{a_1})$ = $P(m_{a_2})$ = $P(m_{a_3})$ = ...

vsathiam wrote: Wait I think the interpretation of the problem is this: Let P(m) denote the product of the nonzero digits of m, and $m_i$ = $P(m_{i-1})$ + $m_{i-1}$, where n = $m_0$. I think you're trying to prove that over all i $\in \mathbb{N}$, there's some infinitely long sequence $a_1, a_2, \dots$ such that $P(m_{a_1})$ = $P(m_{a_2})$ = $P(m_{a_3})$ = ... Yes, this is the problem.

We can take $C$ such that if $n\geq C$, we have $10^{n-1}\geq 9^n$ (Actually, $C=22$ works). Then, if $n$($\geq C$) digits number $m$ is written on the blackboard, we can always transform to $n+1$ digits number of the form $\overline{10A}$, where $A$ is $n-1$ digits number (when immediately after left-most digit is carried up). Assume any $n$($\geq C$) digits number $T$ is written on the black board. We can transform $T$ to $T_1 = \overline{10\cdots 00A_1}$, where digits 0 continues at least once after the first digit 1. If $A_1$ has at least $C$ digits, we can transform $A_1$ to $\overline{10\cdots 0A_2}$. Thus, $T_1$ can be transformed to $T_2 = \overline{10\cdots 010\cdots 0A_2}$. Repeating this procedure, we get $T_k = \overline{10\cdots 010\cdots 010\cdots 0\cdots A_k}$, where $A_k$ has digit at most $C$. Finally, we get $T_k$ with $P(T_k)\leq 9^C$. Therefore, integers which is at most $9^C$ is added infinitely many times. Among them, there is an integer which is added infinitely many times.

For each positive integer $n$ define $f(n)$ to be the product of nonzero digits of $n$. Define $g(n)$ to be the number of digits in $n$ which is not equal to $0$ or $1$. Suppose the initial number is $a_0$ then $a_i=a_{i-1}+f(a_{i-1})$. Let $k$ be the smallest integer with $9^k>10^{k-1}$. Let $m$ be a sufficiently large integer. Let $b_j=\sum_{i=j}^m10^i$. For each $j$ let $h(j)$ be the smallest number such that $a_{h(j)}\geq b_j$. It is easy to see that for all $j\geq k$, the first $m-j$ digit of $a_{h(j)-1}$ must be $1$, hence $$f(a_{h(j)-1})\leq 9^{j+1}$$In particular, $$f(a_{h(k)-1})\leq 9^{k+1}$$which is a constant. Notice that we can define $h(k)$ for each $m$. Therefore, there are infinitely many integers $n$ such that $f(a_n)\leq 9^{k+1}$. By pigeonhole principle some integers must be the image of infinitely many integers.

Wow, just wow. Take a positive integer $k$ satisfying $9^{j} < 10^{j-1}$ for all $j \ge k.$ Now take any integer $m \ge k$ such that the integer $N = \sum_{i=k}^m10^i = \overbrace{11 \dots 11}^{m-k+1 \ \text{digits}} \overbrace{00 \dots 00}^{k \ \text{digits}}$ is greater than the initial integer written. Consider the last integer $L$ written less than this number. Let $l$ be the leftmost digit of $L$ that's not $1$ or $0.$ If $l$'s in the $k$ rightmost digits then we know that the product of the digits is going to be less than $9^k.$ Otherwise, $l$ is in the first $m-k+1$ digits, but $L < N,$ so the digit before it has to be $0.$ Then it's not hard to verify that $L$ plus the product of digits of $L$ can't be greater than $N$ since $9^{j} < 10^{j-1}$ for all $j \ge k.$ Therefore, there are infinitely many integers written with a product of digits bounded by $9^k$ which finishes. $\square$

The original answer is too hard.